Answer on Question #57786, Physics / Mechanics | Relativity
A sailor pushes a 100.0 kg crate up a ramp that is 3.00 m high and 5.00 m long onto the deck
of a ship. He exerts a 650.0 N force parallel to the ramp. What is the mechanical advantage of the
ramp? What is the efficiency of the ramp? Your response should include all of your work and a freebody diagram.
Find: ΔF – ? ƞ – ?
Given:
m=100 kg
h=3 m
l=5 m
F=650 N
g=9,8 N/kg
Solution:
Consider the forces, which acting on the crate.
y
x
⃗
𝑁
𝐹
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝐹𝑓𝑟𝑖𝑐𝑡
α
𝑚𝑔
Newton's Second Law:
∑ni=1 ⃗⃗⃗
Fi = ma⃗ (1)
We believe that the body moves in straight lines and uniformly.
Because a⃗ = ⃗0 (2)
Write the vector sum of all forces:
⃗ + ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗ + mg
∑ni=1 ⃗⃗⃗
Fi = F
Ffrict + N
⃗ (3),
⃗ –traction force,
where F
⃗⃗⃗⃗⃗⃗⃗⃗⃗
Ffrict – friction force,
⃗⃗ – reaction force,
N
mg
⃗ – gravity
(2) and (3) in (1):
⃗ + ⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗ + mg
F
Ffrict + N
⃗ = ⃗0 (4)
Find the projection of forces.
OX: F − Ffrict − mg sin α = 0 (5)
OY: N − mg cos α = 0 (6)
Friction force (with a particular approach):
Ffrict = μN (7),
where μ – coefficient of friction (μ < 1)
Of (6)  N = mg cos α (8)
(8) in (7): Ffrict = μmg cos α (9)
Of (5)  F = Ffrict + mg sin α (10)
(9) in (10): F = mg(μ cos α + sin α) (11)
Expression (μ cos α + sin α) < 1 (12)
Of (11) and (12)  F = mg(μ cos α + sin α) < mg (13)
Of (13)  mechanical advantage of the ramp:
∆F = mg − F (14)
Of  ΔF=330 N
Efficiency of the ramp:
ƞ=
Ahelpful
Aspent
× 100% (15),
where Ahelpful – helpful work,
Aspent – spent work
Helpful work: Ahelpful = mgh (16),
Spent work: Aspent = Fl (17)
(16) and (17) in (15):
ƞ=
mgh
Fl
× 100% (18)
Of (18)  ƞ = 90%
Answer:
ΔF=330 N
ƞ=90%
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